Question: $\int^{2/(3\pi)}_{6/\pi}\dfrac{\cos\left(\dfrac1x\right)}{x^2}\,dx\, = $
Solution: Strategy Let's first find the indefinite integral $\int\dfrac{\cos\left(\dfrac1x\right)}{x^2}\,dx\, $. Then we can use the result to compute the definite integral. Finding the indefinite integral To find $\int\dfrac{\cos\left(\dfrac1x\right)}{x^2}\,dx\, $, we can use U-substitution. If we let $ {u=\dfrac{1}{x}}$, then ${du=-\dfrac{1}{x^2} \, dx}$ and ${ \dfrac1{x^2}\,dx=- du}$. So we have: $\begin{aligned}\int\dfrac{\cos\left(\dfrac1x\right)}{x^2}\,dx\,&=\int\cos\left({\dfrac1x}\right)\cdot {\dfrac{1}{x^2}\ dx}\,\\\\\\\\ &=\int\cos( u)\,\cdot {- du}\,\\\\\\\\ &=-\int\cos(u)\,du\\\\\\\\ &=-\sin(u)+C \\\\\\\\ &=-\sin\left(\dfrac{1}{x}\right)+C \end{aligned}$ Evaluating the definite integral Now let's compute the definite integral: $\begin{aligned}\int^{2/(3\pi)}_{6/\pi}\dfrac{\cos\left(\dfrac1x\right)}{x^2}\,dx\,&=-\sin\left(\dfrac{1}{x}\right)\Bigg|^{2/(3\pi)}_{6/\pi}\\\\\\\\ &=-1\left(\sin\left(\dfrac{3\pi}{2}\right)-\sin\left(\dfrac{\pi}{6}\right)\right)\\\\\\\\ &=-1\left(-1-\dfrac12\right)\\\\\\\\ &=\dfrac32\end{aligned}$ [Did we have to find the indefinite integral first?] The answer $\int^{2/(3\pi)}_{6/\pi}\dfrac{\cos\left(\dfrac1x\right)}{x^2}\,dx\, = \dfrac32$